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There are 12 rolls of film in a box, and 4 are defective. Two rolls are to be selected, one after the other, sampling without replacement. What is the probability of selecting two defective rolls?
Group of answer choices

None of the answers is correct.

7/23

12/132

16/144

1/4

12/144


Sagot :

Using the hypergeometric distribution, it is found that the probability is 12/132

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The rolls are chosen from a sample without replacement, which means that the hypergeometric distribution is used to solve this question.

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Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

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  • 12 rolls means that [tex]N = 12[/tex]
  • 4 defective means that [tex]k = 4[/tex]
  • 2 are chosen, which means that [tex]n = 2[/tex]

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What is the probability of selecting two defective rolls?

This is P(X = 2). So

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,12,2,4) = \frac{C_{4,2}*C_{8,0}}{C_{12,2}} = \frac{4 \times 3}{12 \times 11} = \frac{12}{132}[/tex]

12/132 is the probability.

A similar problem is found at https://brainly.com/question/24229905